Download MATH 31 CLASS NOTES XII – Section 5

Geometry, Chapter 6
Section 6.1: Ratios; Properties of Proportions; Geometric Mean
1.
2.
Ratio – A fraction relating two quantities. A ratio can be represented in four ways:
a
a to b
a:b
ab
b
Proportion – A statement that two ratios are equal.
a c
or a : b = c : d where a, b, c and d are called terms.
=
b d
a is the 1st term, b is the 2nd term, c is the 3rd term and d is the 4th term.
a and d (1st and last terms) are called the extremes and b and c (2nd and 3rd terms) are called the means.
3.
Properties of Proportions
a.
Means-Extremes Property: The product of the means equals the product of the extremes.
a c
x x+2
if and only if ad = bc. Example:
=
⇒ 2( x) =
3( x + 2)
=
3
2
b d
b.
Exchange the Extremes
a c
d c
3 6
10 6
⇒
=
= if and only if
= . Example: =
5
10
5 3
b d
b a
c.
Exchange the Means
a c
a b
3 6
⇒
= if and only if
= . Example: =
5 10
b d
c d
3 5
=
6 10
d.
Invert each Ratio
a c
b d
3 6
⇒
= if and only if
= . Example: =
5 10
b d
a c
5 10
=
3 6
NOTE: This last property is very useful when the variable is in the denominator.
Example:
4.
1 3
=
⇒
x 5
x 5
5
=
⇒ x=
1 3
3
Geometric Mean or Mean Proportional of Two Numbers
a.
Definition – b is the geometric mean of a and c if and only if b = ac . Written as a proportion, we
a b
have = . Then b 2 = ac or b = ac . (thus the name “mean proportional”)
b c
b.
Relating a geometric sequence to the geometric mean.
a1 , a2 , a3 , ..., an is a geometric sequence if and only if the ratio of any 2 consecutive terms is
constant.
This is equivalent to saying: The 2nd of any 3 consecutive terms in a geometric sequence is the
geometric mean between the first and third term.
Consider the 3 consecutive terms a4 , a5 , a6 . Here
c.
a6 a5
= , which gives a5 = a4 a6 .
a5 a4
Example: What is the geometric mean of 3 and 12? of 4 and 7? For each of these, write the three
numbers as a geometric sequence.
Geometry, Chapter 6
Section 6.2: Similar Triangles; Requirements for Similarity
1.
Two polygons are similar if and only if two conditions are satisfied:
(a)
(b)
All pairs of corresponding angles are congruent.
All pairs of corresponding sides are proportional.
Similar Triangles
2.
a)
b)
All three pairs of corresponding angles are congruent.
All pairs of corresponding sides in proportion.
E
B
A
3.
Definition of Similar Triangles
∆ABC  ∆DEF if and only if
C
D
F
∠A ≅ ∠D, ∠B ≅ ∠E , ∠C ≅ ∠F
AB BC AC
and = =
DE EF DF
Ways to Establish Two Triangles are Similar
a. AA Similarity - Two angles in one triangle are congruent respectively to two angles of the other
triangle.
b. SAS Similarity – Two sides of one triangle are proportional to two sides of another triangle and the
included angles are congruent.
c. SSS Similarity – Three sides of one triangle are proportional to three sides of another triangle.
d. A Similarity in a Right Triangle – An acute angle of one right triangle is congruent to an acute angle of
another triangle.
e. LL Similarity in a Right Triangle – The legs of one right triangle are proportional respectively to the
legs of another right triangle.
4.
Prove that the diagonals of a trapezoid divide each other into proportional segments.
Given:
A
Prove:
B
E
D
C
Geometry, Chapter 6
Section 6.3: Four theorems involving similar triangles
1.
B
Midsegment Theorem
A line segment that joins the midpoints
of two sides of a triangle is
(i) parallel to the third side and
(ii) is half the length of the third side.
D
⇒
E
A
C
This line segment is called a midsegment of the triangle.
2.
Geometric Mean (Mean Proportional) in a Right Triangle
A
In a right triangle, the altitude to the
hypotenuse is the geometric mean of the
two segments on the hypotenuse.
D
AD CD
=
CD BD
⇒
B
C
3.
DE C AC and DE = 12 AC
Side Splitting Theorem
B
A line parallel to one side of a
triangle divides the other two sides
into proportional segments.
D
If DE C AC , then
E
A
BD BE
.
=
DA EC
C
Proof: Since DE C AC , ∠ BDE ≅ ∠ BAC because corresponding angles are congruent.
∠ B ≅ ∠ B by the reflexive property. So, ∆ BDE ~ ∆ BAC by AA-Similarity. Therefore,
BD BE
. Then
=
BA BC
BA BC
by the Inversion Property of Proportions.
=
BD BE
BD + DA BE + EC
Since BA = BD + DA and BC = BE + EC, we have
by substitution. This proportion
=
BD
BE
DA
EC
BD DA BE EC
DA EC
or 1 +
. Thus
, which
can be written as the equation
=
1+
+
=
+
=
BD
BE
BD BD BE BE
BD BE
BD BE
by inversion becomes
.
=
DA EC
4.
Midquad Theorem – The line segments connecting the
midpoints of the sides of any quadrilateral form a parallelogram.
If E, F, G and H are midpoints of quadrilateral ABCD as shown,
Then EFGH is a parallelogram.
(Be able to prove this. Hint: Draw a diagonal of ABCD and use the
Midsegment Theorem.)
B
G
C
F
H
A
E
D